Does the Limit Comparison Test Work for Divergent Integrals?

[whatlifeforme]

Homework Statement

use limit comparison test.

Homework Equations

$\displaystyle\int_2^∞ {\frac{1}{\sqrt{x^2 - 1}} dx}$

The Attempt at a Solution

I have tried usin 1/x as the comparison function, but when applying the test it
comes out to 0, not an L -> 0 < L < ∞

 

[Dick]

Homework Statement

use limit comparison test.

Homework Equations

$\displaystyle\int_2^∞ {\frac{1}{\sqrt{x^2 - 1}} dx}$

The Attempt at a Solution

I have tried usin 1/x as the comparison function, but when applying the test it
comes out to 0, not an L -> 0 < L < ∞

Show how you got the limit to be L=0. I get L=1.

 

[whatlifeforme]

$\displaystyle limit (x->inf) \frac{1/x}{1/\sqrt{x^2 - 1}}$

$\displaystyle limit (x->inf) \frac{\sqrt{x^2 - 1}}{x}$ = inf/inf

$\displaystyle limit (x->inf) \frac{2x}{\sqrt{x^2 - 1}}$ = inf/inf

$\displaystyle limit (x->inf) \frac{2}{\sqrt[3/2]{x^2 - 1}}$ = 2/inf = 0

 

[Dick]

I you try to use l'Hopital on that you are just going to go in circles until you make a mistake and miss a chain rule, like you did. Use algebra to simplify the limit. sqrt(x^2-1)=x*sqrt(1-1/x^2).

 

[whatlifeforme]

I you try to use l'Hopital on that you are just going to go in circles until you make a mistake and miss a chain rule, like you did. Use algebra to simplify the limit. sqrt(x^2-1)=x*sqrt(1-1/x^2).

i'm sorry I'm lost, and i don't think i left out the chain rule i just didn't include the simplifications in the work above.

 

[Dick]

i'm sorry I'm lost, and i don't think i left out the chain rule i just didn't include the simplifications in the work above.

I don't know what simplifications you made, since you didn't show them, but they aren't right. The first l'Hopital should give you $\frac{x}{\sqrt{x^2 - 1}}$, the next will give $\frac{\sqrt{x^2 - 1}}{x}$. Etc, etc.

 

[whatlifeforme]

how does this look:

1/sqrt(x^2-1) / (1/x)

$\frac{x}{\sqrt{x^2 - 1}}$

lim (x->inf) $\frac{1}{\sqrt{1-(1/x^2)}}$ = 1

$\displaystyle\int_2^∞ {(1/x) dx}$

$ln|x| ^{∞}_{2}$

diverges.

 

[Dick]

how does this look:

1/sqrt(x^2-1) / (1/x)

$\frac{x}{\sqrt{x^2 - 1}}$

lim (x->inf) $\frac{1}{\sqrt{1-(1/x^2)}}$ = 1

$\displaystyle\int_2^∞ {(1/x) dx}$

$ln|x| ^{∞}_{2}$

diverges.

Looks fine. l'Hopital's isn't the best way to handle every limit.